Datasheet

LTC3618
18
3618fc
For more information www.linear.com/LTC3618
2. I
2
R losses are calculated from the resistances of the
internal switches, R
SW
, and external inductor R
L
. In
continuous mode the average output current flowing
through inductor L is “chopped” between the main
switch and the synchronous switch. Thus, the series
resistance looking into the SW pin is a function of both
top and bottom MOSFET R
DS(ON)
and the duty cycle
(DC), as follows:
R
SW
= (R
DS(ON)TOP
)(DC) + (R
DS(ON)BOT
)(1 – DC)
The R
DS(ON)
for both the top and bottom MOSFETs can
be obtained from the Typical Performance Characteristics
curves. To obtain I
2
R losses, simply add R
SW
to R
L
and
multiply the result by the square of the average output
current.
Other losses, including C
IN
and C
OUT
ESR dissipative losses
and inductor core losses, generally account for less than
2% of the total loss.
Thermal Considerations
In most applications, the LTC3618 does not dissipate
much heat due to its high efficiency. However, in ap
-
plications where the LTC3618 is running at high ambient
temperature with low supply voltage and high duty cycles,
such as in dropout, the heat dissipated may exceed the
maximum junction temperature of the
part. If the junction
temperature reaches approximately 160°C, all four power
switches will be turned off and the SW node will become
high impedance.
To prevent the LTC3618 from exceeding the maximum
junction temperature, the user will need to do some ther
-
mal analysis to determine whether the power dissipated
exceeds the maximum junction temperature of the part.
The temperature rise is given by:
T
RISE
= P
D
• θ
JA
where P
D
is the power dissipated by the regulator, and
θ
JA
is the thermal resistance from the junction of the die
to the ambient temperature. The junction temperature,
T
J
, is given by:
T
J
= T
A
+ T
RISE
where T
A
is the ambient temperature.
As an example, consider this case: the LTC3618 is in
dropout at an input voltage of 3.3V with a load current for
each channel of 2A at an ambient temperature of 70°C.
Assuming a 20°C rise in junction temperature, to 90°C,
results in an R
DS(ON)
of 0.086Ω (see the graph in the
Typical Performance Characteristics section). Therefore,
the power dissipated by the part is:
P
D
= (I
1
2
+ I
2
2
) • R
DS(ON)
= 0.69W
For the QFN package, the θ
JA
is 46.9°C/W.
Therefore, the junction temperature
of the regulator op-
erating at 70°C ambient temperature is approximately:
T
J
= 0.69W • 46.9°C/W + 70°C = 102.4°C
Note that for very low input voltage, the junction tem-
perature will
be higher due to increased switch resistance
R
DS(ON)
. It is not recommended to use full load current at
high ambient temperature and low input voltage.
To maximize the thermal performance of the LTC3618,
the exposed pad should be soldered to a ground plane.
See the PC Board Layout Checklist.
Design Example
As a design example, consider using the LTC3618 in an
application with the following specifications:
V
IN
= 3.3V to 5.5V
V
DDQ
= 1.8V
V
TT
= 0.9V
I
OUT1(MAX)
= 3A
I
OUT2(MAX)
= 3A
I
OUT1(MIN)
= 200mA
f = 2.25MHz
First, calculate the timing resistor:
R
RT
=
4• 10
11
Ω • Hz
2.25MHz
= 178k
applicaTions inForMaTion